Let the event that the first digit is 5 be denoted as \(5_1\) and the event that the second digit is not 5 be denoted as \(\overline{5}_2\).

The probability that the first digit is 5 and the second digit is not 5 is:

\(P(5_1 \cap \overline{5}_2) = \frac{3}{9} \times \frac{6}{8} = \frac{1}{4}\)

The probability that the second digit is not 5 is:

\(P(\overline{5}_2) = \frac{1}{4} + \frac{6}{9} \times \frac{5}{8} = \frac{48}{72} = 0.6666\)

The conditional probability that the first digit is 5 given that the second digit is not 5 is:

\(P(5_1 | \overline{5}_2) = \frac{\frac{1}{4}}{\frac{48}{72}} = \frac{3}{8} = 0.375\)

(i) The probability that Mohit goes to the cinema is 0.35. The probability that he spends less than $50 at the cinema is 1 - 0.8 = 0.2. Therefore, the probability that Mohit will go to the cinema and spend less than $50 is:

\(P(C \cap <50) = 0.35 \times 0.2 = 0.07\)

(ii) We need to find \(P(C \mid <50)\), the probability that Mohit went to the cinema given that he spent less than $50. This is given by:

\(P(C \mid <50) = \frac{P(C \cap <50)}{P(<50)}\)

We already found \(P(C \cap <50) = 0.07\).

Now, calculate \(P(<50)\):

\(P(<50) = P(S \cap <50) + P(C \cap <50) + P(H \cap <50)\)

\(P(S \cap <50) = 0.25 \times (1 - 0.7) = 0.25 \times 0.3 = 0.075\)

\(P(C \cap <50) = 0.07\)

\(P(H \cap <50) = 0.4 \times 1 = 0.4\)

Thus, \(P(<50) = 0.075 + 0.07 + 0.4 = 0.545\)

Therefore, \(P(C \mid <50) = \frac{0.07}{0.545} = 0.128\) (or \(\frac{14}{109}\))

Let \(M\) be the event that Mumbok is chosen, and \(18{-}60\) be the event that a person aged between 18 and 60 is chosen.

The probability of choosing Mumbok and a person aged 18 to 60 is:

\(P(M \cap 18{-}60) = 0.6 \times \frac{55}{90} = 0.367 \text{ or } \frac{11}{30}\)

The probability of choosing a person aged 18 to 60 from either town is:

\(P(18{-}60) = 0.6 \times \frac{55}{90} + 0.4 \times \frac{95}{160} = \frac{29}{48} \text{ or } 0.604\)

Using Bayes' theorem, the probability that the town chosen was Mumbok given that the person is aged 18 to 60 is:

\(P(M \mid 18{-}60) = \frac{P(M \cap 18{-}60)}{P(18{-}60)} = \frac{\frac{11}{30}}{\frac{29}{48}} = \frac{88}{145} \text{ or } 0.607\)

Let \(W_1\) be the event that John wins his first game and \(W_2\) be the event that he wins his second game.

The probability that John wins his second game, \(P(W_2)\), can be calculated as:

\(P(W_2) = P(W_1 \cap W_2) + P(L_1 \cap W_2)\)

\(= 0.3 \times 0.6 + 0.7 \times 0.15\)

\(= 0.18 + 0.105\)

\(= 0.285\)

We need to find \(P(W_1 | W_2)\), the probability that John won his first game given that he won his second game. This is given by:

\(P(W_1 | W_2) = \frac{P(W_1 \cap W_2)}{P(W_2)}\)

\(= \frac{0.18}{0.285}\)

\(= 0.632\)

Thus, the probability that John won his first game given that he won his second game is \(0.632\) or \(\frac{12}{19}\).

(a) The probability that Tom removes a red disc on his first turn can be calculated by considering two scenarios: Sam removes a red disc first, then Tom removes a red disc, or Sam removes a white disc first, then Tom removes a red disc.

Probability of Sam removing a red disc first: \(\frac{3}{8}\)

Probability of Tom removing a red disc after Sam: \(\frac{2}{7}\)

Probability of Sam removing a white disc first: \(\frac{5}{8}\)

Probability of Tom removing a red disc after Sam: \(\frac{3}{7}\)

Total probability: \(\frac{3}{8} \times \frac{2}{7} + \frac{5}{8} \times \frac{3}{7} = \frac{21}{56} = \frac{3}{8}\)

(b) Tom wins on his second turn if the sequence of draws is either RRWR, WRRR, or WRWR.

Probability of RRWR: \(\frac{3}{8} \times \frac{2}{7} \times \frac{5}{6} \times \frac{1}{5} = \frac{1}{56}\)

Probability of WRRR: \(\frac{5}{8} \times \frac{3}{7} \times \frac{2}{6} \times \frac{1}{5} = \frac{1}{56}\)

Probability of WRWR: \(\frac{5}{8} \times \frac{3}{7} \times \frac{2}{6} \times \frac{3}{5} = \frac{1}{14}\)

Total probability: \(\frac{1}{56} + \frac{1}{56} + \frac{1}{14} = \frac{3}{28}\)

(c) Given that Tom wins on his second turn, we need to find the probability that Sam removes a red disc on his first turn.

Probability of Sam removing a red disc first: \(\frac{3}{8}\)

Probability of Tom winning on his second turn given Sam removes a red disc first: \(\frac{1}{56}\)

Total probability of Tom winning on his second turn: \(\frac{3}{28}\)

Conditional probability: \(\frac{\frac{1}{56}}{\frac{3}{28}} = \frac{1}{6}\)